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Chapter 14: Problem 46

When a person exercises, muscle contractions produce lactic acid. Moderate increases in lactic acid can be handled by the blood buffers without decreasing the pH of blood. However, excessive amounts of lactic acid can overload the blood buffer system, resulting in a lowering of the blood pH. A condition called acidosis is diagnosed if the blood pH falls to 7.35 or lower. Assume the primary blood buffer system is the carbonate buffer system described in Exercise \(45 .\) Calculate what happens to the \(\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right] /\left[\mathrm{HCO}_{3}^{-}\right]\) ratio in blood when the \(\mathrm{pH}\) decreases from 7.40 to 7.35.

Short Answer

Expert verified
The ratio of bicarbonate ions (\( \text{HCO}_3^-\)) to carbonic acid (\(\text{H}_2\text{CO}_3\)) increases by a factor of approximately 1.12 when the blood pH decreases from 7.40 to 7.35 due to acidosis.

Step by step solution

01

Write the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is given by: \[ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] In our case, the acid is \(\text{H}_2\text{CO}_3\) (carbonic acid) and the conjugate base is \(\text{HCO}_3^-\) (bicarbonate ion). Let's rewrite the equation: \[ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right) \]
02

Calculate the Change in pH

According to the exercise, the decrease in pH goes from 7.40 to 7.35. The change in pH can be calculated as: \[ \Delta \text{pH} = 7.40 - 7.35 = 0.05 \]
03

Find the \(pK_a\)

The primary blood buffer system is the carbonate buffer system. The \(K_a\) value for carbonic acid is \(4.47 \times 10^{-7}\). In order to use the Henderson-Hasselbalch equation, we need the p\(K_a\) value. To find this, we can use the following equation: \[ \text{p}K_a = -\log_{10}(K_a) \] Plugging in the values, we get: \[ \text{p}K_a = -\log_{10}(4.47 \times 10^{-7}) \approx 6.35\]
04

Calculate the change in the ratio of HCO\(_3^-\) to H\(_2\)CO\(_3\)

Now that we have the change in pH, and the p\(K_a\), we can rearrange the Henderson-Hasselbalch equation to solve for the change in the ratio of bicarbonate ions (\( \text{HCO}_3^-\)) to carbonic acid (\(\text{H}_2\text{CO}_3\)): \[ \Delta \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right) = \Delta \text{pH} - \Delta \text{p}K_a \] Since the p\(K_a\) remains constant (it does not change), the change in p\(K_a\) is zero: \[ \Delta \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right) = 0.05 \] To calculate the change in ratio, we should find the antilog of this value: \[ 10^{\Delta \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right)} = 10^{0.05} \approx 1.12\]
05

Conclusion

The ratio of bicarbonate ions (\( \text{HCO}_3^-\)) to carbonic acid (\(\text{H}_2\text{CO}_3\)) increases by a factor of approximately 1.12 when the blood pH decreases from 7.40 to 7.35 due to acidosis.

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Most popular questions from this chapter

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intracellular fluids at \(\mathrm{pH}\) values generally between 7.1 and 7.2 a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) in intracellular fluid at \(\mathrm{pH}=7.15 ?\) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8}\) b. Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the \(\mathrm{pH}\) of intracellular fluid? \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=7.5 \times 10^{-3}\)

What volume of \(0.0100 \mathrm{M}\) NaOH must be added to \(1.00 \mathrm{L}\) of \(0.0500 \mathrm{M}\) HOCI to achieve a pH of \(8.00 ?\)

Repeat the procedure in Exercise \(61,\) but for the titration of \(25.0 \mathrm{mL}\) of 0.100 \(\mathrm{M}\) \(\mathrm{HNO}_{3}\) with 0.100 \(\mathrm{M}\) \(\mathrm{NaOH}\).

Acid-base indicators mark the end point of titrations by "magically" turning a different color. Explain the "magic" behind acid-base indicators.

A certain buffer is made by dissolving \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\).
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